Question

Find the zeroes of the polynomial x2-6x-27 and verify the relationship between the zeroes and their coefficient ?
${x }^{2} - 6x - 27$

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Answer 1

Given that,

Polynomial, $x^2-6x-27$

To find,

The zeroes of the polynomial.

Solution,

$x^2-6x-27=x^2+3x-9x-27\\\\=x(x+3)-9(x+3)\\\\=(x-9)(x+3)$

For zeroes,

$(x-9)(x+3)=0\\\\x=9, x=-3$

So, the zeros of given polynomial are 9 and -3.

Let $\alpha\ \text{and}\ \beta$ are polynomial were the roots, then :

$\alpha+\beta =\dfrac{-b}{a}=\dfrac{-6}{1}=-6\\\\\alpha+\beta =-6\\\\\text{also}\\\\\alpha \beta =\dfrac{c}{a}=\dfrac{-27}{1}=-27\\\\\alpha \beta=-27$

Hence, this is the required solution.

Answer 2

Step-by-step explanation:

SOLUTION

We are given polynomial x^2-6x-27

We need to find the roots/ zeroes of this polynomial.

To find the zeroes of this Quadratic equation, we need to split the middle term.

On splitting it ,

x^2-6x-27

x^2 - 9x + 3x -27 =0

x(x-9) + 3(x-9) = 0

Hence the zeroes are -3,9